**Previous section: ****Introduction to Geometric Algebra and Basic 2D Applications****.**

__Next section__**: ****Rotors and Rotations in the Euclidean Plane****.**

The algebraic properties of vector addition and scalar multiplication are insufficient to characterize the geometric concept of a vector as a directed line segment, because they fail to encode the properties of magnitude and relative direction. That deficiency is corrected by defining suitable algebraic rules for multiplying vectors.

**Geometric product: **The product

**ab**for

**vectors**

**a, b, c**is defined by the rules

• **Associative: **(**ab**)**c **=** a**(**bc**)

• **Left distributive: ****a**(**b **+ **c**) =** ab + ac**

• **Right distributive: **(**b **+ **c**)**a **=** ba + ca **

• **Euclidean metric: ****a**^{2} = *a*^{2},

where *a *= |**a**|** **is a positive *scalar** *(= real number) called the *magnitude** *of **a**.

In terms of the *geometric product** ***ab **we can define two other products, a symmetric *inner product*

(1) **a****∙****b **= ½(**ab **+ **ba**) = **b****∙****a**

and an antisymmetric **outer product**

(2) **a****∧****b **= ½(**ab **− **ba**) = − **b****∧****a**

Adding (1) and (2), we obtain the fundamental formula

(3) **ab **= **a****∙****b **+ **a****∧****b **called the * expanded form *for the geometric product.

Our next task is to provide geometric interpretations for these three products.

**Exercise: **For a triangle defined by the vector equation **a **+ **b **= **c**,

derive the standard ** Law of Cosines**:

*a*

^{2}+

*b*

^{2}+ 2

**a**

**∙**

**b**=

*c*

^{2}, and so prove that the inner product

**a**

**∙**

**b**is always scalar-valued.

{Just multiply **c**^{2} = (**a** + **b**) (**a** + **b**) and use the Euclidean metric above and definition (1) above for the inner product.

Notice also that if **a** and **b** are perpendicular, using the Pythagorean Theorem we can conclude **a****∙****b** is zero (and vice versa).

Further, if **a** and **b** are perpendicular, using formulas (1) & (2) above, **a****∧****b** = **ab** = ½(**ab **− **ba**), which means **ab** = − **ba**.}

Therefore, the inner product can be given the usual geometric interpretation as the length of a perpendicular projection of one line segment on the direction of another.

The outer product **a****∧****b **= − **b****∧****a **generates a new kind of geometric quantity called a ** bivector**, that can be interpreted geometrically as

*directed area***in the plane of**

**a**and

**b.**

We have shown that the geometric product interrelates three kinds of algebraic entities: * scalars *(0-vectors),

*(1-vectors), and*

**vectors***(2-vectors) that can be interpreted as geometric objects of different dimension. Geometrically, scalars represent 0-dimensional objects, because they have magnitude and orientation (sign) but no direction. Vectors represent directed line segments, which are 1-dimensional objects. Bivectors represent*

**bivectors****, which are 2-dimensional objects. It may be better to refer to a bivector**

*directed plane segments*

as a ** directed area**, because its magnitude

*B*= |

**B**|

**is the ordinary area of the plane segment and its direction**

represents the plane (and orientation) in which the segment lies, just as a unit vector represents the direction of a line. The shape of the plane segment is not represented by any feature of

**B**, as expressed in the following equivalent geometric depictions for

**B**(with

*counterclockwise*orientation):

__Prove the following:__

Given any non-zero vector **a **in the plane of bivector **B**, one can find a vector **b **such that

** B **= **ba **= –**ab**,

** B**^{2} = −|**B**|^{2} = −*a*^{2}*b*^{2},

** aB **= –**Ba**, that is, **B **anticommutes with every vector in the plane of **B**.

{Looking again at the depictions directly above, it should be clear we can pick a **b** perpendicular to **a**

and also in the plane of **B **so that the rectangle depicting **b****∧****a **has the magnitude and orientation of **B**.

Further, since we've picked **b** to be perpendicular to **a**, by the arguments in the exercise above,

**B **= **b****∧****a = ****ba **= – **ab**. Then, **B**^{2} = (–**ab**) (**ba**) = – a^{2}b^{2} = − *a*^{2}*b*^{2} = −|**B**|^{2} < 0.

Finally, **aB **= – **aab** = − *a*^{2}**b **= − **baa** = – **Ba**.}

Every vector **a **has a multiplicative inverse:

** **that is, geometric algebra makes it possible to divide by vectors.

{Simply multiply **a** on either side by **a**/*a*^{2}. As usual, we must assume **a**≠0.}

**Prove the following theorems about ****the geometric meaning of commutivity and anticommutivity****:**

** ****a****∙****b **= 0 ⇔ **ab **= −**ba**** **** ****Orthogonal vectors anticommute!! {See the exercise above.}**

** **

** ****a **= λ**b **⇔ **a****∧****b **= 0 ⇔ **ab **= **ba**** ****Collinear vectors commute!!**

** **

The problem remains to assign geometric meaning to the quantity **ab **without expanding it into inner and outer products.

__Previous section__**: ****Introduction to Geometric Algebra and Basic 2D Applications****.**

**Next section: ****Rotors and Rotations in the Euclidean Plane****.**